Posted by Anonymous on Wednesday, November 2, 2011 at 4:17pm.
We went through this already:
Posted by Anonymous on Tuesday, November 1, 2011 at 1:53pm.
f(x) = sin^2(x/2)
defined on the interval [ -5.683185, 1.270796].
Take a look at that one for the method.
i cant see how you figure that out
I cant find the interval in which it concaves down, and that's the only question i need.
f = sin^2(x/3)
f' = 2/3 sin(x/3)cos(x/3) = 1/3 sin(2x/3)
f'' = 2/9 cos(2x/3)
f is concave down when f'' < 0
so, where is cos(2x/3) < 0?
When pi/2 < 2x/3 < 3pi/2
or 3pi/4 < x < 9pi/4
Your interval appears to be about (-3pi/2,pi/2), so you may have to adjust your answer a bit to fit the interval.
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